MySQL經典題目

一、常見的SQL面試題：經典50題。

來自於知乎分享https://zhuanlan.zhihu.com/p/38354000（部分題目）

表結構：

學生表：student(學號,學生姓名,出生年月,性別)

 

 

成績表：score(學號,課程號,成績)

 

 

 

課程表：course(課程號,課程名稱,教師號)

教師表：teacher(教師號,教師姓名)

 

 

1. 查詢各科成績最高和最低的分， 以如下的形式顯示：課程號，最高分，最低分（groupby分組聚合）

select courseid,max(score),min(score) from score group by courseid;

 2. 查詢平均成績大於60分學生的學號和平均成績（having）

select stuid,avg(score)
from score
group by stuid
having avg(score) > 60;

3. 查詢同名同性學生名單並統計同名人數

select stuname,count(stuid)
from student
group by stuname
having count(stuid) > 1;

 4. 查詢每門課程的平均成績，結果按平均成績升序排序，平均成績相同時，按課程號降序排列

select courseid,avg(score)
from score
group by courseid 
order by avg(score),courseid desc;

 6. 查詢兩門以上不及格課程的同學的學號及其平均成績

select stuid,avg(score) as avg_score
from score
where score < 60
group by stuid
having count(courseid) >=2;

 7. 查詢所有課程成績小於80分學生的學號、姓名

select student.stuid,student.stuname
from student 
where stuid not in(
select distinct stuid 
from score
where score >80
);

 8. 1990年出生的學生名單（stubirtthdate為date類型）

select stuid,stuname
from student 
where year(stubirthdate) = 1990;

 9.

TOPN問題 查詢各科成績最高的記錄（關聯子查詢）

select *
from score as a
where score = (select max(score) from score as b where a.courseid = b.courseid);

上述關聯子查詢順序

1. select * from score

2. select max(score) from score where courseid = 0001

3. select * from score where score =81 and courseid = 0001 

 TOPN問題 查詢各科成績前兩名的記錄（union）

(select * from score where courseid = 0001 order by score desc limit 2)
union
(select * from score where courseid = 0002 order by score desc limit 2)
union
(select * from score where courseid = 0003 order by score desc limit 2);

 10. 查詢出每門課程的及格人數和不及格人數（case表達式）

select courseid,
sum(case when score >= 60 then 1 else 0 end) as jigerenshu,
sum(case when score < 60 then 1 else 0 end) as bujigerenshu 
from score
group by courseid;

11. 使用分段[100-85],[85-70],[70-60],[<60]來統計各科成績，分別統計：各分數段人數，課程號和課程名稱（case表達式）

select score.courseid,course.coursename,
sum(case when score between 85 and 100 then 1 else 0 end) as '[85-100]',
sum(case when score between 70 and 85 then 1 else 0 end) as '[70-85]',
sum(case when score between 60 and 70 then 1 else 0 end) as '[60-70]'
from score right join course
on score.courseid = course.courseid
group by courseid;

 12. 行列互換

參考https://mp.weixin.qq.com/s/6Kll4Q6Xp37i2PiLUh4cMA

下面是學生的成績表（表名score，列名：學號、課程號、成績）

 

使用sql實現將該表行轉列為下面的表結構

 

select stuid,
max(case courseid when '0001' then score else 0 end) as '課程號0001',
max(case courseid when '0002' then score else 0 end) as '課程號0002',
max(case courseid when '0003' then score else 0 end) as '課程號0003'
from score
group by stuid;

 

二、互聯網校招SQL筆試經典50題及答案解析

來自知乎分享https://zhuanlan.zhihu.com/p/53302593（部分題目）

表結構

學生表Student

 

 

 

教師表Teacher

 

 

 

課程表Course

 

成績表SC

 

 

 

1. 查詢“01”課程比“02”課程成績高的所有學生的學號（生成新表 表連接）

select distinct t1.sid as sid
from 
(select * from sc where cid='01') as t1
left join 
(select * from sc where cid='02') as t2
on t1.sid=t2.sid
where t1.score>t2.score;

 2. 查詢學過編號“01”並且也學過編號“02”課程的同學的學號、姓名

select sid,sname
from student
where sid IN
(select `sid`
from SC
group by `sid`
#分組保證 學生 01 和 02課程都學過
having count(if(cid='01',score,null))>0 and count(if(cid='02',score,null))>0);

 3. 查詢所有課程成績小於60分的同學的學號、姓名；

select s_sid.`sid`,student.sname
from (select distinct `sid`
from sc
group by `sid`
having max(score) < 60)s_sid
left join student on s_sid.`sid` = student.`sid`;

 4. 查詢沒有學全所有課的同學的學號、姓名；

select ssid.`sid`,student.sname
from (select `sid`
from sc
group by `sid` having count(`score`) < (select count(*) from course))ssid
left join student on ssid.`sid` = student.`sid`	;

 5. 查詢至少有一門課與學號為“01”的同學所學相同的同學的學號和姓名

select
distinct sc.sid
from 
(
select
cid
from sc
where sid='01'
)t1
left join sc
on t1.cid=sc.cid;

 6. 查詢和"01"號的同學學習的課程完全相同的其他同學的學號和姓名

select t1.`sid` ,student.sname
from
(select sc.`sid`
from 
(
select cid
from sc
where `sid` = 01
)t1
left join sc on t1.`cid` = sc.`cid`
group by `sid`
having count(distinct sc.cid)= (select count(distinct cid) from sc where sid = '01'))t1
left join student on t1.`sid` = student.`sid`;

 7. 查詢兩門及其以上不及格課程的同學的學號，姓名及其平均成績

select ssid.`sid`,student.sname,ssid.`avg(score)`
from 
(
select sid,avg(score)
from sc
group by sid
having count(if(score<60,score,NULL)) >= 2
)ssid left join student on ssid.`sid` = student.`sid`;

 8. 查詢所有課程的成績第2名到第3名的學生信息及該課程成績（窗口函數）

select `sid`,rank_num,score,`cid`
from 
(
select rank() over(PARTITION by `cid` order by score desc) as rank_num,
`sid`,`score`,`cid`
from sc
)t
where rank_num in (2,3);

 9. 統計各科成績各分數段人數：課程編號,課程名稱,[100-85],[85-70],[70-60],[0-60]及所占百分比

select sc.`cid`,cname,
count(if(score between 85 and 100,sid,NULL))/count(sid)
,count(if(score between 70 and 85,sid,null))/count(sid)
,count(if(score between 60 and 70,sid,null))/count(sid)
,count(if(score between 0 and 60,sid,null))/count(sid)
from sc
left join course on sc.`cid` = course.`cid`
group by sc.`cid`;

 10. 查詢學生平均成績及其名次（窗口函數）

select sid,avg_score,rank() over(order by avg_score desc)
from 
(select sid,avg(score) as avg_score
from sc
group by `sid`)t;

 11. 查詢各科成績前三名的記錄（窗口函數）

select * from 
(select `cid`,`sid`,rank() over (PARTITION by `cid` order by score desc) as rank_num
from sc)t
where rank_num <=3;

12. 查詢各學生的年齡 （curdate()）

select sid,sname,year(curdate())-year(sage) as age
from student;

 

2020.4.19 15:00